Given two number X and N, the task is to find the last digit of X raised to last digit of N factorial, i.e. 
.
Examples:
Input: X = 5, N = 2
Output: 5
Explanation:
Since, 2! mod 10 = 2
therefore 52 = 25 and the last digit of 25 is 5.
Input: X = 10, N = 4
Output: 0
Explanation:
Since, 4! mod 10 = 24 mod 10 = 4
therefore 104 = 10000 and the last digit of 10000 is 0.
Approach: The most efficient way to solve this problem is to find any pattern in the required last digit, with the help of last digit of N! and last digit of X raised to Y
Below is the various observation of the above-given equation:
- If N = 0 or N = 1, then the last digit is 1 or
respectively. - Since 5! is 120, therefore for N ? 5 the value of (N! mod 10) will be zero.
- Now we are left with digit 2, 3, 4. For this we have:
for N = 2,
N! mod 10 = 2! mod 10 = 2
for N = 3,
N! mod 10 = 3! mod 10 = 6
for N = 4,
N! mod 10 = 4! mod 10 = 24 mod 10 = 4
Now for X2, X4, and X6
we will check that after which nth power of Xn the value of last digit repeats,
i.e, after which nth power of last digit of Xn the value of last digit repeats.
- Below is the table for what power of the last digit from 0 to 9 in any number repeats:
| Number | Cyclicity |
|---|---|
| 0 | 1 |
| 1 | 1 |
| 2 | 4 |
| 3 | 4 |
| 4 | 2 |
| 5 | 1 |
| 6 | 1 |
| 7 | 4 |
| 8 | 4 |
| 9 | 2 |
Below are the steps based on the above observations:
- If X is not a multiple of 10 then divide the evaluated value of by cyclicity of the last digit of X. If remainder(say r) is 0 then do the following:
- If the last digit of X is any of 2, 4, 6, or 8 then the answer will be 6.
- If the last digit of X is any of 1, 3, 7, or 9 then the answer will be 1.
- If the last digit of X is 5 then answer will be 5.
- Else if remainder(say r) is a non-zero then answer is
, where ‘l’ is the last digit of X. - Else if X is a multiple of 10 then the answer will be 0 always.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include<bits/stdc++.h>using namespace std;// Function to find a^b using // binary exponentiation long power(long a, long b, long c) { // Initialise result long result = 1; while (b > 0) { // If b is odd then, // multiply result by a if ((b & 1) == 1) { result = (result * a) % c; } // b must be even now // Change b to b/2 b /= 2; // Change a = a^2 a = (a * a) % c; } return result; } // Function to find the last digit // of the given equation long calculate(long X, long N) { int a[10]; // To store cyclicity int cyclicity[11]; // Store cyclicity from 1 - 10 cyclicity[1] = 1; cyclicity[2] = 4; cyclicity[3] = 4; cyclicity[4] = 2; cyclicity[5] = 1; cyclicity[6] = 1; cyclicity[7] = 4; cyclicity[8] = 4; cyclicity[9] = 2; cyclicity[10] = 1; // Observation 1 if (N == 0 || N == 1) { return (X % 10); } // Observation 3 else if (N == 2 || N == 3 || N == 4) { long temp = (long)1e18; // To store the last digits // of factorial 2, 3, and 4 a[2] = 2; a[3] = 6; a[4] = 4; // Find the last digit of X long v = X % 10; // Step 1 if (v != 0) { int u = cyclicity[(int)v]; // Divide a[N] by cyclicity // of v int r = a[(int)N] % u; // If remainder is 0 if (r == 0) { // Step 1.1 if (v == 2 || v == 4 || v == 6 || v == 8) { return 6; } // Step 1.2 else if (v == 5) { return 5; } // Step 1.3 else if (v == 1 || v == 3 || v == 7 || v == 9) { return 1; } } // If r is non-zero, // then return (l^r) % 10 else { return (power(v, r, temp) % 10); } } // Else return 0 else { return 0; } } // Else return 1 return 1; } // Driver Code int main(){ // Given Numbers int X = 18; int N = 4; // Function Call long result = calculate(X, N); // Print the result cout << result;} // This code is contributed by spp____ |
Java
import java.util.*;class TestClass { // Function to find a^b using // binary exponentiation public static long power(long a, long b, long c) { // Initialise result long result = 1; while (b > 0) { // If b is odd then, // multiply result by a if ((b & 1) == 1) { result = (result * a) % c; } // b must be even now // Change b to b/2 b /= 2; // Change a = a^2 a = (a * a) % c; } return result; } // Function to find the last digit // of the given equation public static long calculate(long X, long N) { int a[] = new int[10]; // To store cyclicity int cyclicity[] = new int[11]; // Store cyclicity from 1 - 10 cyclicity[1] = 1; cyclicity[2] = 4; cyclicity[3] = 4; cyclicity[4] = 2; cyclicity[5] = 1; cyclicity[6] = 1; cyclicity[7] = 4; cyclicity[8] = 4; cyclicity[9] = 2; cyclicity[10] = 1; // Observation 1 if (N == 0 || N == 1) { return (X % 10); } // Observation 3 else if (N == 2 || N == 3 || N == 4) { long temp = (long) 1e18; // To store the last digits // of factorial 2, 3, and 4 a[2] = 2; a[3] = 6; a[4] = 4; // Find the last digit of X long v = X % 10; // Step 1 if (v != 0) { int u = cyclicity[(int) v]; // Divide a[N] by cyclicity // of v int r = a[(int) N] % u; // If remainder is 0 if (r == 0) { // Step 1.1 if (v == 2 || v == 4 || v == 6 || v == 8) { return 6; } // Step 1.2 else if (v == 5) { return 5; } // Step 1.3 else if (v == 1 || v == 3 || v == 7 || v == 9) { return 1; } } // If r is non-zero, // then return (l^r) % 10 else { return (power(v, r, temp) % 10); } } // Else return 0 else { return 0; } } // Else return 1 return 1; } // Driver's Code public static void main(String args[]) throws Exception { // Given Numbers int X = 18; int N = 4; // Function Call long result = calculate(X, N); // Print the result System.out.println(result); }} |
Python3
# Python3 program for the above approach # Function to find a^b using # binary exponentiation def power(a, b, c): # Initialise result result = 1 while (b > 0): # If b is odd then, # multiply result by a if ((b & 1) == 1): result = (result * a) % c # b must be even now # Change b to b/2 b //= 2 # Change a = a^2 a = (a * a) % c return result# Function to find the last digit # of the given equation def calculate(X, N): a = 10 * [0] # To store cyclicity cyclicity = 11 * [0] # Store cyclicity from 1 - 10 cyclicity[1] = 1 cyclicity[2] = 4 cyclicity[3] = 4 cyclicity[4] = 2 cyclicity[5] = 1 cyclicity[6] = 1 cyclicity[7] = 4 cyclicity[8] = 4 cyclicity[9] = 2 cyclicity[10] = 1 # Observation 1 if (N == 0 or N == 1): return (X % 10) # Observation 3 elif (N == 2 or N == 3 or N == 4): temp = 1e18; # To store the last digits # of factorial 2, 3, and 4 a[2] = 2 a[3] = 6 a[4] = 4 # Find the last digit of X v = X % 10 # Step 1 if (v != 0): u = cyclicity[v] # Divide a[N] by cyclicity # of v r = a[N] % u # If remainder is 0 if (r == 0): # Step 1.1 if (v == 2 or v == 4 or v == 6 or v == 8): return 6 # Step 1.2 elif (v == 5): return 5 # Step 1.3 elif (v == 1 or v == 3 or v == 7 or v == 9): return 1 # If r is non-zero, # then return (l^r) % 10 else: return (power(v, r, temp) % 10) # Else return 0 else: return 0 # Else return 1 return 1# Driver Code if __name__ == "__main__": # Given numbers X = 18 N = 4 # Function call result = calculate(X, N) # Print the result print(result)# This code is contributed by chitranayal |
C#
// C# program for the above approach using System;using System.Collections.Generic;class GFG{ // Function to find a^b using // binary exponentiation static long power(long a, long b, long c) { // Initialise result long result = 1; while (b > 0) { // If b is odd then, // multiply result by a if ((b & 1) == 1) { result = (result * a) % c; } // b must be even now // Change b to b/2 b /= 2; // Change a = a^2 a = (a * a) % c; } return result; } // Function to find the last digit // of the given equation public static long calculate(long X, long N) { int[] a = new int[10]; // To store cyclicity int[] cyclicity = new int[11]; // Store cyclicity from 1 - 10 cyclicity[1] = 1; cyclicity[2] = 4; cyclicity[3] = 4; cyclicity[4] = 2; cyclicity[5] = 1; cyclicity[6] = 1; cyclicity[7] = 4; cyclicity[8] = 4; cyclicity[9] = 2; cyclicity[10] = 1; // Observation 1 if (N == 0 || N == 1) { return (X % 10); } // Observation 3 else if (N == 2 || N == 3 || N == 4) { long temp = (long)1e18; // To store the last digits // of factorial 2, 3, and 4 a[2] = 2; a[3] = 6; a[4] = 4; // Find the last digit of X long v = X % 10; // Step 1 if (v != 0) { int u = cyclicity[(int)v]; // Divide a[N] by cyclicity // of v int r = a[(int)N] % u; // If remainder is 0 if (r == 0) { // Step 1.1 if (v == 2 || v == 4 || v == 6 || v == 8) { return 6; } // Step 1.2 else if (v == 5) { return 5; } // Step 1.3 else if ( v == 1 || v == 3 || v == 7 || v == 9) { return 1; } } // If r is non-zero, // then return (l^r) % 10 else { return (power(v, r, temp) % 10); } } // Else return 0 else { return 0; } } // Else return 1 return 1; } // Driver codestatic void Main() { // Given numbers int X = 18; int N = 4; // Function call long result = calculate(X, N); // Print the result Console.Write(result);}}// This code is contributed by divyeshrabadiya07 |
Javascript
<script> // JavaScript program for the above approach // Function to find a^b using // binary exponentiation function power(a, b, c) { // Initialise result var result = 1; while (b > 0) { // If b is odd then, // multiply result by a if ((b & 1) == 1) { result = (result * a) % c; } // b must be even now // Change b to b/2 b /= 2; // Change a = a^2 a = (a * a) % c; } return result; } // Function to find the last digit // of the given equation function calculate(X, N) { var a = [...Array(10)]; // To store cyclicity var cyclicity = [...Array(11)]; // Store cyclicity from 1 - 10 cyclicity[1] = 1; cyclicity[2] = 4; cyclicity[3] = 4; cyclicity[4] = 2; cyclicity[5] = 1; cyclicity[6] = 1; cyclicity[7] = 4; cyclicity[8] = 4; cyclicity[9] = 2; cyclicity[10] = 1; // Observation 1 if (N == 0 || N == 1) { return X % 10; } // Observation 3 else if (N == 2 || N == 3 || N == 4) { var temp = 1e18; // To store the last digits // of factorial 2, 3, and 4 a[2] = 2; a[3] = 6; a[4] = 4; // Find the last digit of X var v = X % 10; // Step 1 if (v != 0) { var u = cyclicity[parseInt(v)]; // Divide a[N] by cyclicity // of v var r = a[parseInt(N)] % u; // If remainder is 0 if (r == 0) { // Step 1.1 if (v == 2 || v == 4 || v == 6 || v == 8) { return 6; } // Step 1.2 else if (v == 5) { return 5; } // Step 1.3 else if (v == 1 || v == 3 || v == 7 || v == 9) { return 1; } } // If r is non-zero, // then return (l^r) % 10 else { return power(v, r, temp) % 10; } } // Else return 0 else { return 0; } } // Else return 1 return 1; } // Driver Code // Given Numbers var X = 18; var N = 4; // Function Call var result = calculate(X, N); // Print the result document.write(result); // This code is contributed by rdtank. </script> |
Output:
6
Time Complexity: O(logn) because it is using a while loop
Auxiliary Space: O(11)
Approach#2: using for loop
We can first calculate the last digit of N factorial and then take the last digit of X raised to this last digit. We can use modular arithmetic to calculate the last digit of N factorial and to take the last digit of X raised to this value.
Algorithm
1. Calculate the last digit of N factorial using modular arithmetic.
2. Calculate the last digit of X raised to this last digit using modular arithmetic.
3. Return the result.
C++
#include <cmath>#include <iostream>using namespace std;// This function calculates the last digit of X^N factorialint lastDigit(int X, int N){ // Initialize the last digit of N factorial as 1 int lastDigitNFactorial = 1; // Calculate the last digit of N factorial for (int i = 2; i <= N; i++) { lastDigitNFactorial = (lastDigitNFactorial * (i % 10)) % 10; } // Calculate the last digit of X^(N factorial) int lastDigitXToLastDigitNFactorial = (int)pow(X % 10, lastDigitNFactorial) % 10; // Return the result return lastDigitXToLastDigitNFactorial;}int main(){ // Initialize X and N int X = 18; int N = 4; // Print the result of lastDigit(X, N) cout << lastDigit(X, N) << endl; return 0;} |
Java
public class Main { public static int lastDigit(int X, int N) { int lastDigitNFactorial = 1; for (int i = 2; i <= N; i++) { lastDigitNFactorial = (lastDigitNFactorial * (i % 10)) % 10; } int lastDigitXToLastDigitNFactorial = (int)Math.pow(X % 10, lastDigitNFactorial) % 10; return lastDigitXToLastDigitNFactorial; } public static void main(String[] args) { int X = 18; int N = 4; System.out.println(lastDigit(X, N)); }} |
Python3
def last_digit(X, N): last_digit_n_factorial = 1 for i in range(2, N+1): last_digit_n_factorial = (last_digit_n_factorial * (i % 10)) % 10 last_digit_X_to_last_digit_n_factorial = pow( X % 10, last_digit_n_factorial, 10) return last_digit_X_to_last_digit_n_factorialX = 18N = 4print(last_digit(X, N)) |
C#
using System;class Program { // This function calculates the last digit of X^N // factorial static int lastDigit(int X, int N) { // Initialize the last digit of N factorial as 1 int lastDigitNFactorial = 1; // Calculate the last digit of N factorial for (int i = 2; i <= N; i++) { lastDigitNFactorial = (lastDigitNFactorial * (i % 10)) % 10; } // Calculate the last digit of X^(N factorial) int lastDigitXToLastDigitNFactorial = (int)Math.Pow(X % 10, lastDigitNFactorial) % 10; // Return the result return lastDigitXToLastDigitNFactorial; } static void Main(string[] args) { // Initialize X and N int X = 18; int N = 4; // Print the result of lastDigit(X, N) Console.WriteLine(lastDigit(X, N)); }} |
Javascript
// Javascript program implementation function last_digit(X, N) { let last_digit_n_factorial = 1; for (let i = 2; i <= N; i++) { last_digit_n_factorial = (last_digit_n_factorial * (i % 10)) % 10; } let last_digit_X_to_last_digit_n_factorial = Math.pow(X % 10, last_digit_n_factorial) % 10; return last_digit_X_to_last_digit_n_factorial;}let X = 18;let N = 4;console.log(last_digit(X, N)); |
Output
6
Time complexity: O(N), where N is the input
Space complexity: O(1)
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